This question already has an answer here:

Suppose I set up something like in this diagram (sorry for the terrible hand drawing):

enter image description here

When I touch both ends of the cables, will I feel the current/die immediately (if the voltage is high enough)?


marked as duplicate by tpg2114, AccidentalFourierTransform, Gert, Michael Seifert, user36790 Dec 10 '16 at 16:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ No, it is not so bad. With what did you make the drawing? $\endgroup$ – peterh Dec 10 '16 at 2:13
  • $\begingroup$ @peterh Thanks! I used this online tool: sumopaint.com/home and my mouse. $\endgroup$ – YoTengoUnLCD Dec 12 '16 at 23:49
  • $\begingroup$ Note, you don't need a 100000km engineering project for that in the space. You can do this in a room, for around some thousand \$. $\endgroup$ – peterh Dec 13 '16 at 3:35

If the wires are perfect conductors and you connect the terminals sufficiently in the past from when you touch the edges, then yes: there is a slightly higher density of electrons in the one wire, leading to a higher voltage, and the electrons will travel from the higher density to the lower density without realizing that they have to "come from" this battery that's far away.

Of course your conductor might not propagate electric signals quite at the speed of light and it may in theory be feasible to get on a spaceship and outrun the signal that is raising the voltage, touching the opposite side of the wires without getting hurt. Two parallel lines like that form a sort of big RC-circuit with capacitors and resistors in series and parallel, so it does take some time to "fill up" the extra electron density on the one side. Conversely it takes also some time to "drain off" the extra electrons through you, and therefore this assumption I'm making above that $R \to 0$ is crucial for getting the maximum "oomph" on you from the "capacitor" that's in front of you.

  • $\begingroup$ While both effects you mention are important, I think the transmission line effects (the RC-circuit) will dominate on the distances the OP is considering. @YoTengoUnLCD, if you want to explore this question further, I recommend starting with these effects. They're also a nice place to start because there are many examples in communications where these effects are a driving factor, so there's lots of people out there ready to explain them! $\endgroup$ – Cort Ammon Dec 9 '16 at 0:00
  • $\begingroup$ Two very long parallel lines (100.000km long), like envisioned by the OP, charged to a voltage V long time ago, in the initial time after adding a load to the terminals, don't behave like an RC-circuity. See my expanded answer below. $\endgroup$ – freecharly Dec 9 '16 at 19:54

If you have a battery with a voltage of several volts (e.g.12V), you will not feel any current when touching the ends of the cables with your hands no matter how long your cables are. The voltage is simple to low. You might, however, feel something with your tongue, which is more sensitive.

If you use an amperemeter at the end of a very long charged cable, you will measure a current $I=V/Z_0$ immediately corresponding to the cables characteristic impedance $Z_0$ (typically 50-100 $\Omega$) and the battery voltage $V$.

Two very long parallel lines (100.000km long), like envisioned by the OP, charged to a voltage $V$ long time ago, don't behave like an RC-circuity. Given the characteristic impedance $Z_0$, after connecting the terminals to an electric load (like your hands) they act like a voltage source of voltage $V$ with an internal resistance $Z_0$ until the voltage step pulse reaches the battery, is reflected there, and comes back to the terminals. This takes a time, longer (but comparable) to $t=\frac{2·100000km}{300000km/s}=\frac{2}{3} s$ given by the speed of light $c=300000km/s$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.